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Sladkaya [172]
2 years ago
10

1. Based on the observed performance of the air bag models and the amounts of sodium bicarbonate and acetic acid (vinegar) neede

d for an automotive air bag of 80 or 160 L, are these reactants a good substitute for sodium azide? One additional note regarding sodium azide: the rate of inflation after a triggering impact is 40 milliseconds (0.04 s).
Chemistry
1 answer:
Ira Lisetskai [31]2 years ago
7 0

Sodium bicarbonate and acetic acid are not good substitute for sodium azide in airbags since the require more mass and produce less gas.

<h3>Which is the better chemical for an airbag?</h3>

The chemical equation for the production of nitrogen gas from sodium azide is given below:

  • NaN₃ → 2 Na + 3 N₂

1 mole or 66 go of sodium azide produces 3 moles or 67.2 L of nitrogen gas.

The equation for the production of carbon dioxide from sodium bicarbonate and acetic acid is given below:

  • Na₂CO₃ + CH₃COOH → CH₃COONa + CO₂ + H₂O

1 mole, 106 g of Na₂CO₃ and 1 mole, 82 g of CH₃COOH are required to produce 1 mole or 22.4 L of CO₂.

The mass of sodium azide required is less than that of sodium bicarbonate and acetic acid required. Also, sodium azide produces a greater volume of gas. Therefore, sodium bicarbonate and acetic acid are not good substitute for sodium azide in airbags.

In conclusion, sodium azide is a better choice in airbags.

Learn more about airbags at: brainly.com/question/14954949

#SPJ1

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CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7
padilas [110]

<u>Answer:</u> The \Delta G for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)

We are given:

\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol

To calculate \Delta G^o_{rxn} for the reaction, we use the equation:

\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]

For the given equation:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J

Conversion factor used = 1 kJ = 1000 J

The expression of K_p for the given reaction:

K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}

We are given:

p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm

Putting values in above equation, we get:

K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85

To calculate the gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol

Hence, the \Delta G for the reaction is 54.425 kJ/mol

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How many milliliters of a 0.285 M HCl solution are needed to neutralize 249 mL of a 0.0443 M Ba(OH)2 solution?
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Answer:

\large \boxed{\text{77.4 mL}}

Explanation:

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    V/mL:     249

c/mol·L⁻¹:  0.0443     0.285

1. Calculate the moles of Ba(OH)₂

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2. Calculate the moles of HCl

The molar ratio is 2 mol HCl:1 mol Ba(OH)₂

\text{Moles of HCl} = \text{0.011 03 mol Ba(OH)}_{2} \times \dfrac{\text{2 mol HCl}}{\text{1  mol Ba(OH)}_{2}} = \text{0.022 06 mol HCl}

3. Calculate the volume of HCl

V_{\text{HCl}} = \text{0.022 06 mol HCl} \times \dfrac{\text{1 L HCl}}{\text{0.285 mol HCl}} = \text{0.0774 L HCl} = \textbf{77.4 mL HCl}\\\\\text{You must add $\large \boxed{\textbf{77.4 mL}}$ of HCl.}

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